If the solutions of $X'=AX$ have a constant norm then $A$ is skew symmetric. I have a differential equation $X'=AX$ where $A\in\mathcal M_n(\Bbb R)$. The question is to prove that if all the solutions have a constant norm then $A$ is skew-symmetric matrix.
What I have tried so far: let $\varphi(t)=\Vert X(t)\Vert^2$ where $X$ is a solution. Then
$$0=\varphi'(t)=2\langle X'(t),X(t)\rangle=2\langle AX(t),X(t)\rangle=2\langle X(t),A^TX(t)\rangle$$
How can I complete my answer? Any other suggestion?
 A: Now conclude from $x^TAx=0$ for all $x$ that $A_{ii}=0$ for all $i$ by setting $x=e_i$ and then $A_{ij}+A_{ji}=0$ by setting $x=e_i+e_j$.
A: Think that 
$$
\dot X = X \wedge \vec v = A X \mbox{ with } A \ \ \mbox{skew symmetric}
$$
then
$$
< X, \dot X > = < X, X \wedge \vec v > = 0 \Rightarrow ||X|| = C_0
$$
NOTE
The external product can be generalized to $n$ dimensions.
A: The desired result is a consequence of the following
Observation:  A symmetric matrix
$B \in \mathcal M_n(\Bbb R) \tag 1$
such that
$\langle y, By \rangle = 0 \tag 2$
for all vectors $y$ vanishes, i.e.,
$B = 0.\tag 3$
Proof of Observation:
Suppose that
$B \in \mathcal M_n(\Bbb R) \tag{4}$
is symmetric,
$B^T = B; \tag{5}$
then there exists an orthogonal matrix
$O \in \mathcal M_n(\Bbb R), \tag{6}$
that is,
$O^TO = OO^T = I, \tag{7}$
which diagonalizes $B$:
$OBO^T = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag{8}$
where the $\mu_i$, $1 \le i \le n = \text{size}(A)$ are the (real) eigenvalues of $B$.  Now consider the $n \times 1$ (column) vectors
$e_i = (\delta_{ij})_{j = 1}^n, \tag{9}$
that is, $i$-th row entry of $e_i$ is $1$ and all other entries are $0$; we easily see that (8) implies
$(OBO^T)e_i = \mu_i e_i, \tag{10}$
and thus,
$\langle O^Te_i, BO^Te_i \rangle = \langle e_i, OBO^Te_i \rangle = \langle e_i, \mu_i e_i \rangle = \mu_i \langle e_i, e_i \rangle = \mu_i; \tag{11}$
setting
$y_i = O^T e_i, \tag{12}$
we have
$\langle y_i, By_i \rangle = \mu_i; \tag{13}$
now if for all vectors $y$,
$\langle y, By \rangle = 0, \tag{14}$
then in accord with (13),
$\mu_i = 0, \; 1 \le i \le n, \tag{15}$
and thus, via (8)
$B = O^T \text{diag}(\mu_1, \mu_2, \ldots, \mu_n) O = 0. \tag{16}$
Thus a symmetric matrix such that
$\langle y, By \rangle = 0 \tag{17}$
for all vectors $y$ must itself be the $0$ matrix.  End: Proof of Observation.
We apply this Observation to the problem at hand as follows:
with
$\langle x, x \rangle = \text{constant}, \tag{18}$
we have
$\langle x, x \rangle' = 0, \tag{19}$
whence

$\langle \dot x, x \rangle + \langle x, \dot x \rangle = \langle x, x \rangle' = 0; \tag{20}$
given that
$\dot x = Ax, \tag{21}$
we write (20) in the form
$\langle Ax, x \rangle + \langle x, Ax \rangle = 0; \tag{22}$
whence
$\langle x, A^Tx \rangle + \langle x, Ax \rangle = 0, \tag{23}$
or
$\langle x, A^Tx + Ax \rangle = 0, \tag{24}$
or
$\langle x, (A^T + A)x \rangle = 0; \tag{25}$
now $A + A^T$ is symmetric:
$(A + A^T)^T = A^T + (A^T)^T = A^T + A = A + A^T, \tag{26}$
and since $x$ may be any solution to (21), it may be assumed that $x(t)$ may be any vector at time $t$, and hence (25) yields
$\langle x(t), (A^T(t) + A(t))x(t) \rangle = 0; \tag{27}$
we now invoke the above Observation and conclude that
$A^T(t) + A(t) = 0, \tag{28}$
that is,
$A^T(t) = -A(t) \tag{29}$
for all values of $t$.
